2sin^2x+5sinx-5=0 При [-2пи;3пи] Помогите даме решить:( Большое спасибо:*

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Veskrim

13 янв. 2014 г., 10:05:17 (10 лет назад)

2sin²x+5sinx-3=0

Замена sinx = у

2y²+5y-3=0

D = 25 + 24 = 49

x₁ = (-5 -7):4 = -3

x₂ = (-5 + 7):4 = 0.5

sinx = -3 не является решением, т.к. Е(sin x) = [-1,+1]

sinx = 0.5

x = (-1)^k ·π/6 + πk

[-2π;3π] или [-12π/6;18π/6]

k = -2,    x = π/6 - 2π = -11π/6

k = -1,    x = -π/6 - π = -7π/6

k = 0,    x = π/6

k = 1,    x = -π/6 + π = 5π/6

k = 2,    x = π/6 + 2π = 13π/6

k = 3,    x = -π/6 + 3π = 17π/6

Ответ: х = -11π/6, -7π/6, π/6, 5π/6, 13π/6, 17π/6